====== Question 12, Exercise 2.1 ======
Solutions of Question 12 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 12(i)=====
Let $A=\begin{bmatrix}3 & 2 & 1  \\4 & 5 & 6  \\-2 & 3 & 4\end{bmatrix}$. Verify that$A+A^t$ is symmetric.
====Solution====
$$A=\left[ \begin{matrix}
   3 & 2 & 1  \\
   4 & 5 & 6  \\
   -2 & 3 & 4  \\
\end{matrix} \right]$$
$$A^t=\left[ \begin{matrix}
   3 & 4 & -2  \\
   2 & 5 & 3  \\
   1 & 6 & 4  \\
\end{matrix} \right]$$
For symmetric, we have,
$$( A+A^t )^t=A+A^t$$
$$A+A^t=\left[ \begin{matrix}
   3 & 2 & 1  \\
   4 & 5 & 6  \\
   -2 & 3 & 4  \\
\end{matrix} \right]+\left[ \begin{matrix}
   3 & 4 & -2  \\
   2 & 5 & 3  \\
   1 & 6 & 4  \\
\end{matrix} \right]$$
$$=\left[ \begin{matrix}
   3+3 & 2+4 & 1-2  \\
   4+2 & 5+5 & 6+3  \\
   -2+1 & 3+6 & 4+4  \\
\end{matrix} \right]$$
$$A+A^t=\left[ \begin{matrix}
   6 & 6 & -1  \\
   6 & 10 & 9  \\
   -1 & 9 & 8  \\
\end{matrix} \right]$$
$$( A+A^t )^t=\left[ \begin{matrix}
   6 & 6 & -1  \\
   6 & 10 & 9  \\
   -1 & 9 & 8  \\
\end{matrix} \right]$$
$$( A+A^t )^t=( A+A^t )$$

=====Question 12(ii)=====
Let $A=\begin{bmatrix}3 & 2 & 1 \\ 4 & 5 & 6  \\-2 & 3 & 4\end{bmatrix}$. Verify that$A-A^t$ is skew symmetric.
====Solution====
$$A=\left[ \begin{matrix}
   3 & 2 & 1  \\
   4 & 5 & 6  \\
   -2 & 3 & 4  \\
\end{matrix} \right]$$
$$A^t=\left[ \begin{matrix}
   3 & 4 & -2  \\
   2 & 5 & 3  \\
   1 & 6 & 4  \\
\end{matrix} \right]$$
For skew-symmetric, we have,
$$( A-A^t )^t=-( A-A^t)$$
$$A-A^t=\left[ \begin{matrix}
   3 & 2 & 1  \\
   4 & 5 & 6  \\
   -2 & 3 & 4  \\
\end{matrix} \right]-\left[ \begin{matrix}
   3 & 4 & -2  \\
   2 & 5 & 3  \\
   1 & 6 & 4  \\
\end{matrix} \right]$$
$$=\left[ \begin{matrix}
   3-3 & 2-4 & 1+2  \\
   4-2 & 5-5 & 6-3  \\
   -2-1 & 3-6 & 4-4  \\
\end{matrix} \right]$$
$$A-A^t=\left[ \begin{matrix}
   0 & -2 & 3  \\
   2 & 0 & 3  \\
   -3 & -3 & 0  \\
\end{matrix} \right]$$
$$( A-A^t )^t=\left[ \begin{matrix}
   0 & 2 & -3  \\
   -2 & 0 & -3  \\
   3 & 3 & 0  \\
\end{matrix} \right]$$
$$( A-A^t )^t=-\left[ \begin{matrix}
   0 & -2 & 3  \\
   2 & 0 & 3  \\
   -3 & -3 & 0  \\
\end{matrix} \right]$$
$$( A-A^t )^t=-( A-A^t)$$

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